What Is the Sound Intensity Outside a Soundproofed Room?

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Homework Help Overview

The discussion revolves around calculating the sound intensity outside a soundproofed room, given that the room is 45.0 dB quieter than the outside environment. The subject area includes concepts from acoustics and logarithmic relationships in sound intensity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between sound intensity and decibels, exploring the formulas for intensity and sound level. Questions arise about substituting values into the dB equation and the interpretation of reference intensities.

Discussion Status

The discussion is active, with participants attempting to clarify the correct application of the formulas and the relationships between the intensities inside and outside the room. Some participants express uncertainty about their calculations and the use of logarithmic functions.

Contextual Notes

There is a noted confusion regarding the reference intensity values and how to properly apply the dB formula in this context. Participants are also reflecting on their understanding of logarithmic functions as they relate to sound intensity.

sailordragonball
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A recording engineer works in a soundproofed room that is 45.0 dB quieter than the outside. If the sound intensity in the room is 1.30e-10 W/(m^2), what is the intensity outside?

I know this intensity formula ...

... I = (P/A) OR I = ( P / [(4)(pi)(r^2)] )

... and for dB ...

... beta = (10 dB)(log (I/Io)... I can't make sense of the 2 equations to know what to do next ... any ideas?
 
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Do I substitute 45 dB for the 10 in the dB equation?
 
The outside sound is the more intense sound. The intensity inside is a fraction of the intensity of the outside sound such that the ratio gives a 45 dB difference. What is the value of beta, including its sign, if you take the outside sound intensity as the reference? What is it if you take the inside intensity as the reference?
 
I think I have to revisit working with logs ... it's been a while ... LOL - I'll get back to you though.
 
I came up with this ...

... 45 = 10 log ( I / 1e-10 ) = 10 log ( 1.3e-12 / 1e-10 ) ...

... does that sound right? LOL
 
sailordragonball said:
I came up with this ...

... 45 = 10 log ( I / 1e-10 ) = 10 log ( 1.3e-12 / 1e-10 ) ...

... does that sound right? LOL

No it doesn't. The 45 is in the right place, but where did 1e-10 come from? With a +45 on the left, the sound intensity in the room should be the reference I_o and the intensity outside the I. what is the inverse function of the log functiuon?
 

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